FP1 June 2010 Q9
9.
Using the standard results for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n} r^2\),
| Scheme | Marks |
|---|---|
| If \(n = 1\), \(\displaystyle\sum_{r=1}^{n} r^2 = 1\) and \(\dfrac{1}{6}n(n + 1)(2n + 1) = \dfrac{1}{6} \times 1 \times 2 \times 3 = 1\), so true for \(n = 1\). | B1 |
| Assume result true for \(n = k\) | M1 |
| \(\displaystyle\sum_{r=1}^{k+1} r^2 = \frac{1}{6}k(k + 1)(2k + 1) + (k + 1)^2\) | M1 |
| \(= \dfrac{1}{6}(k + 1)(2k^2 + 7k + 6)\) or \(= \dfrac{1}{6}(k + 2)(2k^2 + 5k + 3)\) or \(= \dfrac{1}{6}(2k + 3)(k^2 + 3k + 2)\) | A1 |
| \(= \dfrac{1}{6}(k + 1)(k + 2)(2k + 3) \quad = \dfrac{1}{6}(k + 1)\big(\{k + 1\} + 1\big)\big(2\{k + 1\} + 1\big)\) or equivalent | dM1 |
| True for \(n = k + 1\) if true for \(n = k\), (and true for \(n = 1\)) so true by induction for all \(n\). | A1cso |
| (6) |
Alternative for (a) After first three marks B M M1 as earlier:
| Scheme | Marks |
|---|---|
| B1M1M1 | |
| May state RHS \(= \dfrac{1}{6}(k + 1)\big(\{k + 1\} + 1\big)\big(2\{k + 1\} + 1\big) = \dfrac{1}{6}(k + 1)(k + 2)(2k + 3)\) for third M1 | dM1 |
| Expands to \(\dfrac{1}{6}(k + 1)(2k^2 + 7k + 6)\) and show equal to \(\displaystyle\sum_{r=1}^{k+1} r^2 = \frac{1}{6}k(k + 1)(2k + 1) + (k + 1)^2\) for A1 | A1 |
| So true for \(n = k + 1\) if true for \(n = k\), and true for \(n = 1\), so true by induction for all \(n\). | A1cso |
| (6) |
Notes
(a) B1: Checks \(n = 1\) on both sides and states true for \(n = 1\) here or in conclusion
M1: Assumes true for \(n = k\) (should use one of these two words)
M1: Adds \((k+1)\)th term to sum of \(k\) terms
A1: Correct work to support proof
M1: Deduces \(\tfrac{1}{6}n(n + 1)(2n + 1)\) with \(n = k + 1\)
A1: Makes induction statement. Statement true for \(n = 1\) here could contribute to B1 mark earlier
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n}(r^2 + 5r + 6) = \sum_{r=1}^{n} r^2 + 5\sum_{r=1}^{n} r + \left(\sum_{r=1}^{n} 6\right)\) | M1 |
| \(\dfrac{1}{6}n(n + 1)(2n + 1) + \dfrac{5}{2}n(n + 1), \quad + 6n\) | A1, B1 |
| \(= \dfrac{1}{6}n\left[(n + 1)(2n + 1) + 15(n + 1) + 36\right]\) | M1 |
| \(= \dfrac{1}{6}n\left[2n^2 + 18n + 52\right] = \dfrac{1}{3}n(n^2 + 9n + 26)\) or \(a = 9,\ b = 26\) | A1 |
| (5) |
Notes
(b) M1: Expands and splits (but allow 6 rather than sigma 6 for this mark)
A1: first two terms correct
B1: for \(6n\)
M1: Take out factor \(n\)/6 or \(n\)/3 correctly – no errors factorising
A1: for correct factorised cubic or for identifying \(a\) and \(b\)
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=n+1}^{2n}(r + 2)(r + 3) = \frac{1}{3}2n(4n^2 + 18n + 26) - \frac{1}{3}n(n^2 + 9n + 26)\) | M1 A1ft |
| \(\dfrac{1}{3}n(8n^2 + 36n + 52 - n^2 - 9n - 26) = \dfrac{1}{3}n(7n^2 + 27n + 26)\) (*) | A1cso |
| (3) | |
| 14 marks |
Notes
(c) M1: Try to use \(\displaystyle\sum_{1}^{2n}(r + 2)(r + 3) - \sum_{1}^{n}(r + 2)(r + 3)\) with previous result used at least once
A1ft Two correct expressions for their \(a\) and \(b\) values
A1: Completely correct work to printed answer